**Solution:**

Let the first player be moving from left to right. By convention left to right is taken as the positive direction and thus

right to left is the negative direction . If symbols m and u represent the mass and initial velocity of the two

players, respectively. Subscripts 1 and 2 in these physical quantities refer to the two hockey players. Thus,

m_{1} = 60 kg; u_{1} = + 5 m s^{-1}; and m_{2 }= 55 kg; u_{2 }= – 6 m s^{-1}

.

The total momentum of the two players before the collision

= 60 kg × (+ 5 m s^{-1}) + 55 kg × (– 6 m s^{-1})

= – 30 kg m s^{-1} If v is the velocity of the two entangled players after the collision, the total momentum then

= (m_{1} + m_{2 }) × v = (60 + 55) kg × v m s–1

= 115 × v kg m s^{–1}

.

Equating the momenta of the system before and after collision, in accordance with the law of conservation of

momentum, we get

v = – 30/115 = – 0.26 m s^{–1}

.

Thus, the two entangled players would move with velocity 0.26 m s^{–1} from right to left, that is, in the direction the

second player was moving before the collision.